Question

the four vertices of a parallelogram are plotted on the coordinate plane shown above. what is the distance, in units, between vertices a and d?

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Identify coordinates

Assume the coordinates of point $A$ are $(x_1,y_1)$ and of point $D$ are $(x_2,y_2)$ from the coordinate - plane.

Step3: Substitute into formula

Calculate $\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Since the coordinates of $A$ and $D$ are not given in the text (but would be read from the graph), let's assume $A=(x_1,y_1)$ and $D=(x_2,y_2)$. If $A=(a,b)$ and $D=(c,d)$, then the distance $d=\sqrt{(c - a)^2+(d - b)^2}$. For example, if $A=(1,1)$ and $D=(4,5)$:

Step4: Calculate values

$(c - a)=4 - 1=3$ and $(d - b)=5 - 1 = 4$. Then $(c - a)^2+(d - b)^2=3^2+4^2=9 + 16=25$.

Step5: Find square - root

$d=\sqrt{25}=5$.

Answer:

The distance depends on the actual coordinates of $A$ and $D$ read from the graph. If we assume $A=(x_1,y_1)$ and $D=(x_2,y_2)$, the distance is $\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. If specific coordinates are provided (e.g., $A=(1,1)$ and $D=(4,5)$), the distance is 5.